D - Pagodas

Posted 2020-12-24 wzl19981116

n pagodas were standing erect in Hong Jue Si between the Niushou Mountain and the Yuntai Mountain, labelled from 1 to n. However, only two of them (labelled a and b, where 1≤a≠b≤n) withstood the test of time.
Two monks, Yuwgna and Iaka, decide to make glories great again. They take turns to build pagodas and Yuwgna takes first. For each turn, one can rebuild a new pagodas labelled i (i?{a,b} and 1≤i≤n) if there exist two pagodas standing erect, labelled j and k respectively, such that i=j+k or i=j?k. Each pagoda can not be rebuilt twice.
This is a game for them. The monk who can not rebuild a new pagoda will lose the game.

Input

The first line contains an integer t (1≤t≤500) which is the number of test cases.
For each test case, the first line provides the positive integer n (2≤n≤20000) and two different integers a and b.

Output

For each test case, output the winner (Yuwgna" orIaka"). Both of them will make the best possible decision each time.

Sample Input

16
2 1 2
3 1 3
67 1 2
100 1 2
8 6 8
9 6 8
10 6 8
11 6 8
12 6 8
13 6 8
14 6 8
15 6 8
16 6 8
1314 6 8
1994 1 13
1994 7 12

Sample Output

Case #1: Iaka
Case #2: Yuwgna
Case #3: Yuwgna
Case #4: Iaka
Case #5: Iaka
Case #6: Iaka
Case #7: Yuwgna
Case #8: Yuwgna
Case #9: Iaka
Case #10: Iaka
Case #11: Yuwgna
Case #12: Yuwgna
Case #13: Iaka
Case #14: Yuwgna
Case #15: Iaka
Case #16: Iaka

自己设几个数模拟一下，可以发现，他们只能相差两个人的最大公约数，然后看看有奇数个还是偶数个就可以了

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<cmath>
#include<string.h>
#include<algorithm>
#define sf scanf
#define pf printf
#define pb push_back
#define mm(a,b) memset((a),(b),sizeof(a))
#include<vector>
typedef __int64 ll;
typedef long double ld;
const ll mod=1e9+7;
using namespace std;
const double pi=acos(-1.0);
int gcd(int a,int b)
{
    return b==0?a:gcd(b,a%b);
 }
bool judge(int m,int a,int b)
{
        int x=gcd(a,b);
        int num=m/x;
        if(num&1)
            return false;
        else
            return true;
}
int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        int m,a,b;
        sf("%d%d%d",&m,&a,&b);
        if(judge(m,a,b))
        pf("Case #%d: Iaka
",i);
        else
        pf("Case #%d: Yuwgna
",i); 
    }
    return 0;
}